Irodov 3.37: Spheroids, Cones and Hyperboloids – Amit & Deepali Sinha Foundation

Problem Irodov 3.37

An electrostatic potential is $\varphi = a(x^2 + y^2) + bz^2$ , with $a$ and $b$ constants. Find the electric field vector, its magnitude and its direction. Describe the shape of the equipotential surfaces when

$a > 0$ and $b > 0$ ;
$a > 0$ and $b < 0$ .

Given

$\varphi = a(x^2 + y^2) + bz^2$
(a) $a > 0$ , $b > 0$
(b) $a > 0$ , $b < 0$

Find

\vect{E}

, its magnitude and direction; the shape of the equipotentials in (a) and (b)

Solution

Step 1The field as minus the gradient

The field is minus the gradient of the potential, component by component:

\begin{aligned} E_x &= -\frac{\partial\varphi}{\partial x} = -2ax\\ E_y &= -\frac{\partial\varphi}{\partial y} = -2ay\\ E_z &= -\frac{\partial\varphi}{\partial z} = -2bz \end{aligned}

\vect{E} = -2\left(ax\,\vect{i} + ay\,\vect{j} + bz\,\vect{k}\right)

(1)

Its magnitude is the root of the sum of the squared components:

E = 2\sqrt{a^2(x^2 + y^2) + b^2z^2}

(2)

For $a > 0$ the part of $\vect{E}$ across the $z$ axis points toward that axis; its part along the axis points toward the plane $z = 0$ when $b > 0$ and away from it when $b < 0$ .

Step 2The equipotentials: a surface of revolution

The potential depends on $x$ and $y$ only through $\rho^2 = x^2 + y^2$ , the squared distance from the $z$ axis, so every equipotential is a surface of revolution about that axis. Its shape is that of its section in a plane through the axis, the curve

a\rho^2 + bz^2 = \varphi_0 \tag{3}

for the level $\varphi_0$ .

Step 3Case (a): both constants positive

With $a, b > 0$ the left side of (3) is never negative, so only levels $\varphi_0 > 0$ exist, and dividing by $\varphi_0$ puts the section in the standard form of an ellipse:

\frac{\rho^2}{\varphi_0/a} + \frac{z^2}{\varphi_0/b} = 1

\begin{gathered}\text{(a) ellipsoids of revolution about } z\\ \text{semi-axes } \sqrt{\varphi_0/a} \text{ and } \sqrt{\varphi_0/b}\end{gathered}

(4)

The semi-axis $\sqrt{\varphi_0/a}$ lies across the $z$ axis and $\sqrt{\varphi_0/b}$ along it. Every field line runs inward to the origin, where $\varphi$ is least.

Case (a) in the plane y = 0, for a = 2b > 0: the elliptical sections of the equipotentials (teal), elongated along z, and the field lines (crimson) running into the origin. — Case (a) in the plane $y = 0$ , for $a = 2b > 0$ : the elliptical sections of the equipotentials (teal), elongated along $z$ , and the field lines (crimson) running into the origin.

Step 4Case (b): constants of opposite sign

With $a > 0$ and $b < 0$ , write $b = -\lvert b\rvert$ ; the section (3) becomes

a\rho^2 - \lvert b\rvert z^2 = \varphi_0

whose shape depends on the sign of the level. For $\varphi_0 > 0$ it is a hyperbola crossing the plane $z = 0$ ; for $\varphi_0 < 0$ a hyperbola crossing the $z$ axis, in two branches; and for $\varphi_0 = 0$ the pair of straight lines

\frac{\rho}{\lvert z\rvert} = \sqrt{\frac{\lvert b\rvert}{a}}

which are the asymptotes of all the others. Rotated about the $z$ axis:

\begin{gathered}\text{(b) } \varphi_0 > 0:\ \text{one-sheet hyperboloids}\\ \varphi_0 = 0:\ \text{a circular cone}\\ \varphi_0 < 0:\ \text{two-sheet hyperboloids}\end{gathered}

(5)

All three are surfaces of revolution about the $z$ axis, and the cone is the common asymptote of both families of hyperboloids.

Case (b) in the plane y = 0, for a = -2b > 0: hyperbolic sections (teal) about the two lines of the cone φ = 0; the field lines (crimson) come in along x and leave along z. — Case (b) in the plane $y = 0$ , for $a = -2b > 0$ : hyperbolic sections (teal) about the two lines of the cone $\varphi = 0$ ; the field lines (crimson) come in along $x$ and leave along $z$ .

Check

An electrostatic field must be curl free. From (1), $\partial E_z/\partial x = 0 = \partial E_x/\partial z$ , and likewise for the other pairs, since each component depends only on its own coordinate. On the $z$ axis the magnitude reduces to $2\lvert b\rvert\lvert z\rvert$ , the slope of $bz^2$ alone, and in the plane $z = 0$ to $2a\rho$ , the slope of $a\rho^2$ alone. With $a = b$ the ellipsoids of (a) become spheres and the field the radial $-2a\vect{r}$ .

Remark

The source of this field follows from Poisson’s equation: $\nabla^2\varphi = 4a + 2b = -\rho_{\text{ch}}/\varepsilon_0$ , a uniform charge density $\rho_{\text{ch}} = -2\varepsilon_0(2a + b)$ . Case (a) needs negative charge filling space, and a charge-free region requires $b = -2a$ , which is case (b) with a cone of half-angle $\arctan\sqrt{2} \approx 54.7^\circ$ about the $z$ axis: the potential near the centre of a Paul ion trap.